49be0df8-3ae9-46aa-b36c-a368990c539a.html

<h4>Common Errors When Using the Quadratic Formula</h4>
<p>The quadratic formula has many parts in it. A small error in any one part can lead to incorrect solutions.</p>
<p>Suppose we are solving \(2x^2&minus;6=11x\). To use the formula, let's rewrite it in the form of \(ax^2+bx+c=0\),
  which gives: \(2x^2&minus;11x&minus;6=0\).</p>
<br>
<p>Here are some common errors to avoid:</p>
<p> <strong>Using the wrong values for \(a\), \(b\), and \(c\) in the formula. </strong></p>
<p>\(x=\frac{-b \pm \sqrt{b^2- 4ac}}{2a}\)</p>
<p>\(x=\frac{-11 \pm \sqrt{(11)^2- 4(2)(-6)}}{2(2)}\)</p>
<p>This is incorrect! \(b\) is -11, so \(-b\) is -(-11), which is 11, not -11.</p>
<p>The correct formula is:<br>
  \(x=\frac{11 \pm \sqrt{(11)^2- 4(2)(-6)}}{2(2)}\)</p>
<br>
<p> <strong>Forgetting to multiply 2 by \(a\) for the denominator in the formula. </strong></p>
<p>\(x=\frac{11 \pm \sqrt{(11)^2- 4(2)(-6)}}{2}\)</p>
<p>This is incorrect! The denominator is \(2a\), which is \(2(2)\) or 4.</p>
<p>The correct formula is:<br>
  \(x=\frac{11 \pm \sqrt{(11)^2- 4(2)(-6)}}{2(2)}\)</p>
<br>
<p> <strong>Forgetting that squaring a negative number produces a positive number. </strong></p>
<p>\(x=\frac{11 \pm \sqrt{-121- 4(2)(-6)}}{4}\)</p>
<p>This is incorrect! \((-11)^2\) is 121, not -121.</p>
<p>The correct formula is:<br>
  \(x=\frac{11 \pm \sqrt{121- 4(2)(-6)}}{4}\)</p>
<br>
<p> <strong>Forgetting that a negative number times a positive number is a negative number. </strong></p>
<p>\(x=\frac{11 \pm \sqrt{121- 48}}{4}\)</p>
<p>This is incorrect! \(4(2)(-6)=-48\) and \(121&minus;(-48)\) is \(121+48\).</p>
<p>The correct formula is:<br>
  \(x=\frac {11 \pm \sqrt{121+48}}{4}\)</p>
<br>
<p> <strong>Making calculation errors or not following the properties of algebra.</strong></p>
<p>\(x=\frac {11 \pm \sqrt{169}}{4}\)</p>
<p>\(x=11 \pm \sqrt{42.25}\)</p>
<p>This is incorrect! Both parts of the numerator, the 11 and the \(\sqrt{169}\), get divided by 4. Also,
  \(\frac{\sqrt{169}}{4}\) is not \(\sqrt{42.25}\).</p>
<p>The correct formula is:<br>
  \(x=\frac{11 \pm 13}{4}\)</p>
<p>Let's finish by evaluating \(\frac{11 \pm 13}{4}\) correctly:<br>
  \(x=\frac{11+13}{4}\) or \(x=\frac{11-13}{4}\)<br>
  \(x=\frac{24}{4}\) or \(x=-\frac{2}{4}\)<br>
  \(x=6\) or \(x=-\frac{1}{2}\)</p>
<p>To make sure our solutions are indeed correct, we can substitute the solutions back into the original equations and see whether each solution keeps the equation true.</p>
<p>Checking 6 as a solution:<br>
  \(2x^2-6=11x\)<br>
  \(2(6)^2-6=11(6)\)<br>
  \(2(36)-6=66\)<br>
  \(72-6=66\)<br>
  \(66=66 \; \checkmark\)</p>
<p>Checking \(-\frac{1}{2}\) as a solution:<br>
  \(2x^2-6=11x\)<br>
  \(2(-\frac{1}{2})^2-6=11(-\frac{1}{2})\)<br>
  \(2(\frac{1}{4})-6=-\frac{11}{2}\)<br>
  \(\frac{1}{2}-6=-5\frac{1}{2}\)<br>
  \(-5\frac{1}{2}=-5\frac{1}{2} \; \checkmark\)</p>
<p>We can also graph the equation \(y=2x^2&minus;11x&minus;6\) and find its \(x\)-intercepts to see whether our solutions to \(2x^2&minus;11x&minus;6=0\) are accurate (or close to accurate).</p>
<p><img height="147" src="https://k12.openstax.org/contents/raise/resources/ece2ac2c7be190db20c3f6a08962cb7fc60f742e" width="262"></p>
<h4>Try It: Common Errors When Using the Quadratic Formula</h4>
<div class="os-raise-ib-cta" data-button-text="Solution" data-fire-event="Reveal1" data-schema-version="1.0">
  <div class="os-raise-ib-cta-content">
    <p>Zariah was solving the equation \(3x^2-8x+4=0\) with the quadratic formula and used the steps below. She then had to stop because she realized there was a problem. Locate and describe her error.</p>
    <p><strong>Step 1 -</strong> \(x=\frac{-(-8) \pm \sqrt{-8^2-4(3)(4)}}{6}\)</p>
    <p><strong>Step 2 -</strong> \(x=\frac{8 \pm \sqrt{-64-48}}{6}\)</p>
  </div>
  <div class="os-raise-ib-cta-prompt">
    <p>Write down your answer, then select the <strong>solution</strong> button to compare your work.</p>
  </div>
</div>
<div class="os-raise-ib-content" data-schema-version="1.0" data-wait-for-event="Reveal1">
  <p>Here is how to find the error:</p>
  <p>First, Zariah substituted \(a=3\), \(b= -8\), and \(c =4\) into the formula in Step 1.</p>
  <p>However, under the radical, \(b^2=(-8)^2= + 64\).</p>
  <p>Step 2 would then become:</p>
  <p>\(x=\frac{8 \pm \sqrt{64-48}}{6}\)</p>
  <p> Which leads to the following solution:</p>
  <p>\(x=\frac{8 \pm \sqrt{16}}{6}\)</p>
  <p>\(x=\frac{8 \pm 4}{6}\)</p>
  <p>\(x=\frac{12}{6}=2\), \(x=\frac{4}{6}=\frac{2}{3}\)</p>
</div>